Bài giải:

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

= -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - $\frac{1}{8}$ = (2x)³ – ($\frac{1}{2}$)³ = (2x - $\frac{1}{2}$)[(2x)² + 2x . $\frac{1}{2}$ + ($\frac{1}{2}$)²]

^{= (2x - $\frac{1}{2}$)(4x2 + x + $\frac{1}{4}$)}

d) $\frac{1}{25}$x² – 64y² = ${\left(\frac{1}{5}x\right)}^2$- (8y)² = ($\frac{1}{5}$x + 8y)($\frac{1}{5}$x - 8y)

a) x2 + 6x + 9 = x2 + 2.3x + 32 = (x + 3)2

b) 10x – 25 – x2 = -(x2 -10x + 25) = -(x2 -2.5x + 52)

= -(x – 5)2

c) 8x3 – 1/8= (2x)3 – ( 1/2)3 = (2x – 1/2)[(2x)2 + 2x . 1/2+ (1/2)2]

= (2x – 1/2)(4x2 + x + 1/4)

d) 1/25x2 – 64y2 = (1/5 x)2– (8y)2 = ( 1/5 x + 8y)(1/5x- 8y)

a) \(x^2+6x+9\)

\(=\left(x+3\right)^2\)

\(=\left(x+3\right)\left(x+3\right)\)

b) \(10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

\(=-\left(x-5\right)\left(x-5\right)\)

c) \(8x^3-\frac{1}{8}\)

\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

d) \(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)

b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)

c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)

d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)

\(4x^4+4x^2+1=\left(2x^2+1\right)^2\)

\(9x^4-6x^2+1=\left(3x^2-1\right)^2\)

\(\dfrac{x^2}{9}-\dfrac{2}{3}x+1=\left(\dfrac{x}{3}+1\right)^2\)

\(x^2-25=\left(x-5\right)\left(x+5\right)\)

Lời giải:

a.

$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$

b.

$=\frac{1}{3}x(y+3xz+3z)$

c.

$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$

d.

$=\frac{1}{5}x^2(2+25x+5y)$

a)  \(x^2\)\(+\)\(6x\)\(+\)\(9\)

\(=\left(x+3\right)^2\)

b)  \(x^3\)\(+\)\(3x^2\)\(+\)\(3x\)\(+\)\(1\)

\(=\left(x+1\right)^3\)

c)  \(8x^3\)\(-\)\(\frac{1}{8}\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

d)  \(10x\)\(-\)\(25\)\(-\)\(x^2\)

\(=\)\(-x^2\)\(+\)\(10\)\(-\)\(25\)

\(=-\left(x^2-10+25\right)\)

\(=-\left(x-5\right)^2\)

e)  \(\frac{1}{25}x^2\)\(-\)\(64y^2\)

=\(\left(\frac{1}{25}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

= (x + y - 3x)(x + y + 3x)

= (y - 2x)(4x + y)

e) (x + 5)² - (2x - 1)²

= (x + 5 - 2x + 1)(x + 5 + 2x - 1)

= (6 - x)(3x + 4)

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

(x - 10)(x + 2) = 0

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2

\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)

\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)

a) Kết quả 2x(2x – 3).             b) Kết quả xy( x 2  – 2xy + 5).

c) Kết quả 2x(x + 1)(x + 4).    d) Kết quả 2 5 ( y − 1 ) ( x + y ) .

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)